leetcode392 Is Subsequence 

 1 """
 2  Given a string s and a string t, check if s is subsequence of t.
 3 You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).
 4 A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).
 5 Example 1:
 6 s = "abc", t = "ahbgdc"
 7 Return true.
 8 Example 2:
 9 s = "axc", t = "ahbgdc"
10 Return false.
11 """
12 """
13 自强自信题。
14 看到这道题的想法就是 在子串中遍历每个字符，
15 按顺序判断是否在父串中
16 如果在，flag为True，父串变为所匹配字母的下标+1到末尾
17 如果不在，flag为False，跳出循环
18 """
19 class Solution1(object):
20     def isSubsequence(self, s, t):
21         n = len(s)       #子串长度
22         if n == 0:
23             return True
24         flag = False
25         for i in range(n):
26             if s[i] in t:                #判断字符是否在t里
27                 t = t[t.index(s[i])+1:]  #bug没有加1
28                 flag = True
29             else:
30                 flag = False
31                 break            #bug没有提前跳出，后面的字符如果匹配成功，最后会是True
32         return flag
33 
34 """
35 解法二：提供一种队列法。与Solution1思想一致。
36 这里用队列十分高级
37 """
38 class Solution2(object):
39     def isSubsequence(self, s, t):
40         from collections import deque
41         queue = deque(s)          #deque模块提供了两端都可以操作的序列，
42                                   # 这意味着，在序列的前后你都可以执行添加或删除操作
43         for c in t:
44             if not queue:
45                 return True
46             if c == queue[0]:       # d = deque('12345')
47                 queue.popleft()     # d.popleft()  '1'   extendleft
48         return not queue            # d.pop()      '5'   extend
49 
50 """
51 第三种是种双指针法，很重要，许多题目都可以使用
52 双指针要用while循环，单独操作i, j步长，总长度为终止条件
53 """
54 class Solution3(object):
55     def isSubsequence(self, s, t):
56         slen = len(s)
57         tlen = len(t)
58         i, j = 0, 0                      #bug代码
59         while(i < slen and j < tlen):    #for i in range(slen):
60             if(s[i] == t[j]):            #    for j in range(tlen):
61                 i += 1                   #        if s[i] == t[i]:
62             j += 1
63         return i == slen

 

posted @ 2020-02-05 10:06  yawenw  阅读(125)  评论(0)    收藏  举报